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Unformatted text preview: Chapter 27Current and ResistanceCHAPTE R OUTLI N E27.1 Electric Current27.2 Resistance27.3 A Model for ElectricalConduction27.4 Resistance and Temperature27.5 Superconductors27.6 Electrical PowerL These power lines transfer energy from the power company to homes and businesses.The energy is transferred at a very high voltage, possibly hundreds of thousands of volts insome cases. Despite the fact that this makes power lines very dangerous, the high voltageresults in less loss of power due to resistance in the wires. (Telegraph Colour Library/ FPG)831Thus far our treatment of electrical phenomena has been conned to the study ofcharges in equilibrium situations, or electrostatics. We now consider situations involvingelectric charges that are not in equilibrium. We use the term electric current, or simplycurrent, to describe the rate of ow of charge through some region of space. Most practical applications of electricity deal with electric currents. For example, the battery in aashlight produces a current in the lament of the bulb when the switch is turned on.A variety of home appliances operate on alternating current. In these common situations, current exists in a conductor, such as a copper wire. It also is possible forcurrents to exist outside a conductor. For instance, a beam of electrons in a televisionpicture tube constitutes a current.This chapter begins with the denition of current. A microscopic description of current is given, and some of the factors that contribute to the opposition to the ow ofcharge in conductors are discussed. A classical model is used to describe electrical conduction in metals, and some of the limitations of this model are cited. We also deneelectrical resistance and introduce a new circuit element, the resistor. We conclude bydiscussing the rate at which energy is transferred to a device in an electric circuit.27.1++++A+IFigure 27.1 Charges in motionthrough an area A. The time rate atwhich charge ows through thearea is dened as the current I.The direction of the current is thedirection in which positive chargesow when free to do so.Electric CurrentIn this section, we study the ow of electric charges through a piece of material. Theamount of ow depends on the material through which the charges are passing and thepotential difference across the material. Whenever there is a net ow of charge throughsome region, an electric current is said to exist.It is instructive to draw an analogy between water ow and current. In manylocalities it is common practice to install lowow showerheads in homes as a waterconservation measure. We quantify the ow of water from these and similar devices byspecifying the amount of water that emerges during a given time interval, which isoften measured in liters per minute. On a grander scale, we can characterize a rivercurrent by describing the rate at which the water ows past a particular location. Forexample, the ow over the brink at Niagara Falls is maintained at rates between1 400 m3/s and 2 800 m3/s.There is also an analogy between thermal conduction and current. In Section 20.7,we discussed the ow of energy by heat through a sample of material. The rate ofenergy ow is determined by the material as well as the temperature difference acrossthe material, as described by Equation 20.14.To dene current more precisely, suppose that charges are moving perpendicular toa sur face of area A, as shown in Figure 27.1. (This area could be the crosssectional areaof a wire, for example.) The current is the rate at which charge ows through thissurface. If Q is the amount of charge that passes through this area in a time intervalt, the average current I av is equal to the charge that passes through A per unit time:Iav832Qt(27.1)S E C T I O N 2 7. 1 Electric Current833If the rate at which charge ows varies in time, then the current varies in time; wedene the instantaneous current I as the differential limit of average current:dQdtI(27.2)Electric currentThe SI unit of current is the ampere (A):1A1C1s(27.3)That is, 1 A of current is equivalent to 1 C of charge passing through the sur face areain 1 s.The charges passing through the sur face in Figure 27.1 can be positive or negative,or both. It is conventional to assign to the current the same direction as the owof positive charge. In electrical conductors, such as copper or aluminum, the currentis due to the motion of negatively charged electrons. Therefore, when we speak ofcurrent in an ordinary conductor, the direction of the current is opposite thedirection of ow of electrons. However, if we are considering a beam of positivelycharged protons in an accelerator, the current is in the direction of motion of theprotons. In some casessuch as those involving gases and electrolytes, for instancethe current is the result of the ow of both positive and negative charges.If the ends of a conducting wire are connected to form a loop, all points on theloop are at the same electric potential, and hence the electric eld is zero within andat the sur face of the conductor. Because the electric eld is zero, there is no nettransport of charge through the wire, and therefore there is no current. However, ifthe ends of the conducting wire are connected to a battery, all points on the loop arenot at the same potential. The battery sets up a potential difference between the endsof the loop, creating an electric eld within the wire. The electric eld exerts forces onthe conduction electrons in the wire, causing them to move in the wire, thus creating acurrent.It is common to refer to a moving charge (positive or negative) as a mobile chargecarrier. For example, the mobile charge carriers in a metal are electrons.We can relate current to the motion of the charge carriers by describing a microscopicmodel of conduction in a metal. Consider the current in a conductor of crosssectionalarea A (Fig. 27.2). The volume of a section of the conductor of length x (the grayregion shown in Fig. 27.2) is A x. If n represents the number of mobile chargecarriers per unit volume (in other words, the charge carrier density), the number ofcarriers in the gray section is nA x. Therefore, the total charge Q in this section isnumber of carriers in sectioncharge per carrier(nA x)qwhere q is the charge on each carrier. If the carriers move with a speed vd , the displacement they experience in the x direction in a time interval t is x vd t. Let uschoose t to be the time interval required for the charges in the cylinder to movethrough a displacement whose magnitude is equal to the length of the cylinder. Thistime interval is also that required for all of the charges in the cylinder to pass throughthe circular area at one end. With this choice, we can write Q in the formQ27.1 Current Flow IsRedundantThe phrase current ow is commonly used, although it is strictlyincorrect, because current is aow (of charge). This is similarto the phrase heat transfer, whichis also redundant because heat isa transfer (of energy). We willavoid this phrase and speak ofow of charge or charge ow .xvdAvd tFigure 27.2 A section of a uniformconductor of crosssectional area A.The mobile charge carriers movewith a speed vd , and the displacement they experience in the xdirection in a time interval t isx vd t. If we choose t tobe the time interval during whichthe charges are displaced, onthe average, by the length of thecylinder, the number of carriers inthe section of length x is n Avd t ,where n is the number of carriersper unit volume.(nAvd t)qIf we divide both sides of this equation byconductor isI avPITFALL PR EVE NTI O NqMicroscopic Model of CurrentQLQtt, we see that the average current in thenqvdA(27.4)Current in a conductor in termsof microscopic quantities834C HAPTE R 2 7 Current and ResistancevdEFigure 27.3 A schematicrepresentation of the zigzagmotion of an electron in aconductor. The changes indirection are the result of collisionsbetween the electron and atoms inthe conductor. Note that the netmotion of the electron is oppositethe direction of the electric eld.Because of the acceleration of thecharge carriers due to the electricforce, the paths are actuallyparabolic. However, the drift speedis much smaller than the averagespeed, so the parabolic shape is notvisible on this scale.The speed of the charge carriers vd is an average speed called the drift speed. Tounderstand the meaning of drift speed, consider a conductor in which the charge carriers are free electrons. If the conductor is isolatedthat is, the potential differenceacross it is zerothen these electrons undergo random motion that is analogous to themotion of gas molecules. As we discussed earlier, when a potential difference is appliedacross the conductor (for example, by means of a battery), an electric eld is set up inthe conductor; this eld exerts an electric force on the electrons, producing a current.However, the electrons do not move in straight lines along the conductor. Instead, theycollide repeatedly with the metal atoms, and their resultant motion is complicated andzigzag (Fig. 27.3). Despite the collisions, the electrons move slowly along the conductor (in a direction opposite that of E) at the drift velocity vd .We can think of the atomelectron collisions in a conductor as an effective internalfriction (or drag force) similar to that experienced by the molecules of a liquid owingthrough a pipe stuffed with steel wool. The energy transferred from the electrons tothe metal atoms during collisions causes an increase in the vibrational energy of theatoms and a corresponding increase in the temperature of the conductor.Quick Quiz 27.1Consider positive and negative charges moving horizontally through the four regions shown in Figure 27.4. Rank the current in these fourregions, from lowest to highest.+++++++++(a)(b)(c)(d)Figure 27.4 (Quick Quiz 27.1) Charges move through four regions.Quick Quiz 27.2Electric charge is conserved. As a consequence, when currentarrives at a junction of wires, the charges can take either of two paths out of the junctionand the numerical sum of the currents in the two paths equals the current that enteredthe junction. Thus, current is (a) a vector (b) a scalar (c) neither a vector nor a scalar.Example 27.1Drift Speed in a Copper WireThe 12gauge copper wire in a typical residential buildinghas a crosssectional area of 3.31 10 6 m2. If it carries acurrent of 10.0 A, what is the drift speed of the electrons?Assume that each copper atom contributes one freeelectron to the current. The density of copper is 8.95 g/cm3.Solution From the periodic table of the elements inAppendix C, we nd that the molar mass of copper is63.5 g/mol. Recall that 1 mol of any substance containsAvogadros number of atoms (6.02 1023). Knowing thedensity of copper, we can calculate the volume occupied by63.5 g ( 1 mol) of copper:Vm63.5 g8.95 g/cm37.09 cm3Because each copper atom contributes one free electron tothe current, we have6.021023 electrons7.09 cm38.49n1028 electrons/m31.00106 cm31 m3From Equation 27.4, we nd that the drift speed isInqAvdwhere q is the absolute value of the charge on each electron.Thus,vdInqA(8.492.221028 m10410.0 C/s10 19 C)(3.313)(1.60m/s106m2)S E C T I O N 2 7. 2 ResistanceExample 27.1 shows that typical drift speeds are ver y low. For instance, electronstraveling with a speed of 2.22 10 4 m/s would take about 75 min to travel 1 m! Inview of this, you might wonder why a light turns on almost instantaneously when aswitch is thrown. In a conductor, changes in the electric eld that drives the freeelectrons travel through the conductor with a speed close to that of light. Thus,when you ip on a light switch, electrons already in the lament of the lightbulbexperience electric forces and begin moving after a time inter val on the order ofnanoseconds.27.2 ResistanceIn Chapter 24 we found that the electric eld inside a conductor is zero. However, thisstatement is true only if the conductor is in static equilibrium. The purpose of thissection is to describe what happens when the charges in the conductor are not in equilibrium, in which case there is an electric eld in the conductor.Consider a conductor of crosssectional area A carrying a current I. The currentdensity J in the conductor is dened as the current per unit area. Because the currentI nqvd A, the current density isIAJnqvdL835PITFALL PR EVE NTI O N27.2 Electrons AreAvailable EverywhereElectrons do not have to travel fromthe light switch to the light in orderfor the light to operate. Electronsalready in the lament of thelightbulb move in response to theelectric eld set up by the battery.Notice also that a battery doesnot provide electrons to thecircuit. It establishes the electriceld that exerts a force onelectrons already in the wires andelements of the circuit.(27.5)where J has SI units of A/m2. This expression is valid only if the current density is uniform and only if the sur face of crosssectional area A is perpendicular to the directionof the current. In general, current density is a vector quantity:Jnq vd(27.6)Current densityFrom this equation, we see that current density is in the direction of charge motionfor positive charge carriers and opposite the direction of motion for negativecharge carriers.A current density J and an electric eld E are established in a conductorwhenever a potential difference is maintained across the conductor. In somematerials, the current density is proportional to the electric eld:JE(27.7)where the constant of proportionality is called the conductivity of the conductor.1Materials that obey Equation 27.7 are said to follow Ohms law, named after GeorgSimon Ohm (17891854). More specically, Ohms law states thatfor many materials (including most metals), the ratio of the current density to theelectric eld is a constant that is independent of the electric eld producing thecurrent.Materials that obey Ohms law and hence demonstrate this simple relationshipbetween E and J are said to be ohmic. Experimentally, however, it is found that not allmaterials have this property. Materials and devices that do not obey Ohms law are saidto be nonohmic. Ohms law is not a fundamental law of nature but rather an empiricalrelationship valid only for certain materials.We can obtain an equation useful in practical applications by considering asegment of straight wire of uniform crosssectional area A and length , as shown in1Do not confuse conductivitywith sur face charge density, for which the same symbol is used.Georg Simon OhmGerman physicist (17891854)Ohm, a high school teacher andlater a professor at the Universityof Munich, formulated theconcept of resistance anddiscovered the proportionalitiesexpressed in Equations 27.7 and27.8. ( Bettmann/Corbis)836LC HAPTE R 2 7 Current and ResistancePITFALL PR EVE NTI O N27.3 Weve SeenSomething LikeEquation 27.8 BeforeIn Chapter 5, we introducedNewtons second law, F ma,for a net force on an object ofmass m. This can be written asmFaIn that chapter, we dened massas resistance to a change in motion inresponse to an external force. Mass asresistance to changes in motion isanalogous to electrical resistanceto charge ow, and Equation 27.8is analogous to the form ofNewtons second law shown here.AVbVaEFigure 27.5 A uniform conductor of length and crosssectional area A. A potentialdifference V Vb Va maintained across the conductor sets up an electric eld E,and this eld produces a current I that is proportional to the potential difference.Figure 27.5. A potential difference V Vb Va is maintained across the wire, creating in the wire an electric eld and a current. If the eld is assumed to be uniform, thepotential difference is related to the eld through the relationship2VJMany individuals call Equation27.8 Ohms law, but this is incorrect. This equation is simply thedenition of resistance, and provides an important relationshipbetween voltage, current, andresistance. Ohms law is relatedto a linear relationship between Jand E (Eq. 27.7) or, equivalently,between I and V, which, fromEquation 27.8, indicates that theresistance is constant, independent of the applied voltage.VEI/A, we can write the potential difference asVPITFALL PR EVE NTI O N27.4 Equation 27.8 Is NotOhms LawETherefore, we can express the magnitude of the current density in the wire asBecause JLIJIARIThe quantity R/ A is called the resistance of the conductor. We can dene theresistance as the ratio of the potential difference across a conductor to the current inthe conductor:VIR(27.8)We will use this equation over and over again when studying electric circuits. From thisresult we see that resistance has SI units of volts per ampere. One volt per ampere isdened to be one ohm ( ):1V1A1(27.9)This expression shows that if a potential difference of 1 V across a conductor causes acurrent of 1 A, the resistance of the conductor is 1 . For example, if an electricalappliance connected to a 120V source of potential difference carries a current of 6 A,its resistance is 20 .The inverse of conductivity is resistivity3 :1Resistivity is the inverse ofconductivity(27.10)where has the units ohmmeters ( m). Because Rresistance of a uniform block of material along the lengthResistance of a uniformmaterial along the lengthR2(27.11)AThis result follows from the denition of potential difference:Vb3/ A, we can express theasDo not confuse resistivityVabaE dsE0dxEwith mass density or charge density, for which the same symbol is used.S E C T I O N 2 7. 2 ResistanceLTable 27.1Resistivities and Temperature Coefcients of Resistivityfor Various MaterialsMaterialResistivitya(SilverCopperGoldAluminumTungstenIronPlatinumLeadNichromecCarbonGermaniumSiliconGlassHard rubberSulfurQuartz (fused)1.59 10 81.7 10 82.44 10 82.82 10 85.6 10 810 10 811 10 822 10 81.50 10 63.5 10 50.466401010 to 10141013101575 1016m)TemperatureCoefcientb [( C)3.83.93.43.94.55.03.923.90.40.548751010101010101010101010101]33333333333All values at 20C.bSee Section 27.4.cA nickelchromium alloy commonly used in heating elements.Henry Leap and Jim LehmanEvery ohmic material has a characteristic resistivity that depends on the properties ofthe material and on temperature. Additionally, as you can see from Equation 27.11, theresistance of a sample depends on geometry as well as on resistivity. Table 27.1 givesthe resistivities of a variety of materials at 20C. Note the enormous range, from verylow values for good conductors such as copper and silver, to very high values for goodinsulators such as glass and rubber. An ideal conductor would have zero resistivity, andan ideal insulator would have innite resistivity.Equation 27.11 shows that the resistance of a given cylindrical conductor such as awire is proportional to its length and inversely proportional to its crosssectional area.If the length of a wire is doubled, then its resistance doubles. If its crosssectional areais doubled, then its resistance decreases by one half. The situation is analogous to theow of a liquid through a pipe. As the pipes length is increased, the resistance to owincreases. As the pipes crosssectional area is increased, more liquid crosses a givencross section of the pipe per unit time interval. Thus, more liquid ows for the samepressure differential applied to the pipe, and the resistance to ow decreases.An assortment of resistors used in electrical circuits.PITFALL PR EVE NTI O N27.5 Resistance andResistivity3a837Resistivity is property of a substance, while resistance is a property of an object. We have seensimilar pairs of variables before.For example, density is a property of a substance, while mass isa property of an object. Equation27.11 relates resistance to resistivity, and we have seen a previousequation (Equation 1.1) whichrelates mass to density.838C HAPTE R 2 7 Current and ResistanceTable 27.2Color Coding for ResistorsFigure 27.6 The colored bands on a resistor represent a code fordetermining resistance. The rst two colors give the rst two digits in theresistance value. The third color represents the power of ten for themultiplier of the resistance value. The last color is the tolerance of theresistance value. As an example, the four colors on the circled resistors arered ( 2), black ( 0), orange ( 103), and gold ( 5%), and so theresistance value is 20 10320 k with a tolerance value of 5% 1 k .(The values for the colors are from Table 27.2.)NumberMultiplierBlackBrownRedOrangeYellowGreenBlueVioletGrayWhiteGoldSilverColorlessSuperStockColor01234567891101102103104105106107108109101012Tolerance5%10%20%Most electric circuits use circuit elements called resistors to control the currentlevel in the various parts of the circuit. Two common types of resistors are the composition resistor, which contains carbon, and the wirewound resistor, which consists of a coil ofwire. Values of resistors in ohms are normally indicated by colorcoding, as shown inFigure 27.6 and Table 27.2.Ohmic materials and devices have a linear currentpotential difference relationship over a broad range of applied potential differences (Fig. 27.7a). The slope of theIversus V curve in the linear region yields a value for 1/R. Nonohmic materials havea nonlinear currentpotential difference relationship. One common semiconductingdevice that has nonlinear Iversus V characteristics is the junction diode (Fig. 27.7b).The resistance of this device is low for currents in one direction (positive V ) andhigh for currents in the reverse direction (negative V ). In fact, most modernelectronic devices, such as transistors, have nonlinear currentpotential differencerelationships; their proper operation depends on the particular way in which theyviolate Ohms law.Quick Quiz 27.3Suppose that a currentcarrying ohmic metal wire has acrosssectional area that gradually becomes smaller from one end of the wire to theother. The current must have the same value in each section of the wire so that chargedoes not accumulate at any one point. How do the drift velocity and the resistance perIISlope = 1RV(a)V(b)Figure 27.7 (a) The currentpotential difference curve for an ohmic material. Thecurve is linear, and the slope is equal to the inverse of the resistance of the conductor.(b) A nonlinear currentpotential difference curve for a junction diode. This devicedoes not obey Ohms law.S E C T I O N 2 7. 2 Resistance839unit length vary along the wire as the area becomes smaller? (a) The drift velocity andresistance both increase. (b) The drift velocity and resistance both decrease. (c) Thedrift velocity increases and the resistance decreases. (d) The drift velocity decreasesand the resistance increases.Quick Quiz 27.4A cylindrical wire has a radius r and length . If both r andare doubled, the resistance of the wire (a) increases (b) decreases (c) remains the same.Quick Quiz 27.5 In Figure 27.7b, as the applied voltage increases, the resistance of the diode (a) increases (b) decreases (c) remains the same.Example 27.2The Resistance of a ConductorCalculate the resistance of an aluminum cylinder thathas a length of 10.0 cm and a crosssectional area of2.00 10 4 m2. Repeat the calculation for a cylinder of thesame dimensions and made of glass having a resistivity of3.0 1010m.Solution From Equation 27.11 and Table 27.1, we can calculate the resistance of the aluminum cylinder as follows:R(2.82A1.4110Example 27.3108m)0.100 m2.00 10 4 m25Solution The crosssectional area of this wire isA(0.321A(3.0A1010m)0.100 m2.00 10 4 m210131.5As you might guess from the large difference in resistivities,the resistances of identically shaped cylinders of aluminumand glass differ widely. The resistance of the glass cylinder is18 orders of magnitude greater than that of the aluminumcylinder.Interactive103m)23.24101.5 10 6m7 m23.24 104.6Solution Because a 1.0m length of this wire has a resistance of 4.6 , Equation 27.8 givesI7m2The resistivity of Nichrome is 1.5 10 6m (seeTable 27.1). Thus, we can use Equation 27.11 to nd theresistance per unit length:RRThe Resistance of Nichrome Wire(A) Calculate the resistance per unit length of a 22gaugeNichrome wire, which has a radius of 0.321 mm.r2Similarly, for glass we nd that/m(B) If a potential difference of 10 V is maintained across a1.0m length of the Nichrome wire, what is the current inthe wire?VR10 V4.62.2 ANote from Table 27.1 that the resistivity of Nichrome wire isabout 100 times that of copper. A copper wire of the sameradius would have a resistance per unit length of only0.052 /m. A 1.0m length of copper wire of the sameradius would carry the same current (2.2 A) with an appliedpotential difference of only 0.11 V.Because of its high resistivity and its resistance tooxidation, Nichrome is often used for heating elements intoasters, irons, and electric heaters.Explore the resistance of different materials at the Interactive Worked Example link at http://www.pse6.com.Example 27.4The Radial Resistance of a Coaxial CableCoaxial cables are used extensively for cable television andother electronic applications. A coaxial cable consists oftwo concentric cylindrical conductors. The regionbetween the conductors is completely lled with silicon, asshown in Figure 27.8a, and current leakage through thesilicon, in the radial direction, is unwanted. (The cable isdesigned to conduct current along its lengththis is notthe current we are considering here.) The radius of theinner conductor is a 0.500 cm, the radius of the outerone is b 1.75 cm, and the length is L 15.0 cm.840C HAPTE R 2 7 Current and ResistanceLCur rentdirectiondrSiliconarbInnerconductorOuterconductor(a)End view(b)Figure 27.8 (Example 27.4) A coaxial cable. (a) Silicon lls the gap between the twoconductors. (b) End view, showing current leakage.Calculate the resistance of the silicon between the twoconductors.Solution Conceptualize by imagining two currents, assuggested in the text of the problem. The desired current isalong the cable, carried within the conductors. Theundesired current corresponds to charge leakage throughthe silicon and its direction is radial. Because we know theresistivity and the geometry of the silicon, we categorize thisas a problem in which we nd the resistance of the siliconfrom these parameters, using Equation 27.11. Because thearea through which the charges pass depends on the radialposition, we must use integral calculus to determine theanswer.To analyze the problem, we divide the silicon into concentric elements of innitesimal thickness dr (Fig. 27.8b).We start by using the differential form of Equation 27.11,replacingwith r for the distance variable: dRdr/A,where dR is the resistance of an element of silicon ofthickness dr and sur face area A. In this example, we take asour representative concentric element a hollow silicon cylinder of radius r, thickness dr, and length L, as in Figure 27.8.Any charge that passes from the inner conductor to theouter one must pass radially through this concentric element, and the area through which this charge passes isA 2 rL. (This is the curved sur face areacircumferencemultiplied by lengthof our hollow silicon cylinder ofthickness dr.) Hence, we can write the resistance of ourhollow cylinder of silicon asdRdr2 rLBecause we wish to know the total resistance across theentire thickness of the silicon, we must integrate this expression from r a to r b :(1)RbadRb2Ladrr2LlnbaSubstituting in the values given, and usingsilicon, we obtain640mln2 (0.150 m)R1.75 cm0.500 cm640m for851To nalize this problem, let us compare this resistance tothat of the inner conductor of the cable along the 15.0cmlength. Assuming that the conductor is made of copper, wehaveRA3.2(1.710108m)0.150 m(5.00 10 3 m)25This resistance is much smaller than the radial resistance. Asa consequence, almost all of the current corresponds tocharge moving along the length of the cable, with a verysmall fraction leaking in the radial direction.What If? Suppose the coaxial cable is enlarged to twice theoverall diameter with two possibilities: (1) the ratio b/a isheld xed, or (2) the difference b a is held xed. For whichpossibility does the leakage current between the inner andouter conductors increase when the voltage is appliedbetween the two conductors?Answer In order for the current to increase, the resistancemust decrease. For possibility (1), in which b/a is held xed,Equation (1) tells us that the resistance is unaffected. Forpossibility (2), we do not have an equation involving the difference b a to inspect. Looking at Figure 27.8b, however,we see that increasing b and a while holding the voltageconstant results in charge owing through the samethickness of silicon but through a larger overall area perpendicular to the ow. This larger area will result in lower resistance and a higher current.S E C T I O N 2 7. 3 A Model for Electrical Conduction27.3841A Model for Electrical ConductionIn this section we describe a classical model of electrical conduction in metals that wasrst proposed by Paul Drude (18631906) in 1900. This model leads to Ohms law andshows that resistivity can be related to the motion of electrons in metals. Although theDrude model described here does have limitations, it nevertheless introduces conceptsthat are still applied in more elaborate treatments.Consider a conductor as a regular array of atoms plus a collection of free electrons,which are sometimes called conduction electrons. The conduction electrons, althoughbound to their respective atoms when the atoms are not part of a solid, gain mobilitywhen the free atoms condense into a solid. In the absence of an electric eld, the conduction electrons move in random directions through the conductor with average speeds onthe order of 106 m/s. The situation is similar to the motion of gas molecules conned in avessel. In fact, some scientists refer to conduction electrons in a metal as an electron gas.There is no current in the conductor in the absence of an electric eld because the driftvelocity of the free electrons is zero. That is, on the average, just as many electrons movein one direction as in the opposite direction, and so there is no net ow of charge.This situation changes when an electric eld is applied. Now, in addition to undergoing the random motion just described, the free electrons drift slowly in a directionopposite that of the electric eld, with an average drift speed vd that is much smaller(typically 10 4 m/s) than their average speed between collisions (typically 106 m/s).Figure 27.9 provides a crude description of the motion of free electrons in aconductor. In the absence of an electric eld, there is no net displacement after manycollisions (Fig. 27.9a). An electric eld E modies the random motion and causes theelectrons to drift in a direction opposite that of E (Fig. 27.9b).In our model, we assume that the motion of an electron after a collision is independent of its motion before the collision. We also assume that the excess energy acquiredby the electrons in the electric eld is lost to the atoms of the conductor when theelectrons and atoms collide. The energy given up to the atoms increases their vibrational energy, and this causes the temperature of the conductor to increase. The temperature increase of a conductor due to resistance is utilized in electric toasters andother familiar appliances.We are now in a position to derive an expression for the drift velocity. When a freee) is subjected to an electric eld E, itelectron of mass me and charge q (experiences a force F q E. Because this force is related to the acceleration of theelectron through Newtons second law, F me a, we conclude that the acceleration ofthe electron isqEmea(27.12)E(a)(b)Active Figure 27.9 (a) A schematic diagram of the random motion of two chargecarriers in a conductor in the absence of an electric eld. The drift velocity is zero.(b) The motion of the charge carriers in a conductor in the presence of an electriceld. Note that the random motion is modied by the eld, and the charge carriershave a drift velocity.At the Active Figures linkat http://www.pse6.com, youcan adjust the electric eld tosee the resulting effect on themotion of an electron.842C HAPTE R 2 7 Current and ResistanceThis acceleration, which occurs for only a short time interval between collisions,enables the electron to acquire a small drift velocity. If vi is the electrons initial velocitythe instant after a collision (which occurs at a time that we dene as t 0), then thevelocity of the electron at time t (at which the next collision occurs) isvfviatqEtmevi(27.13)We now take the average value of vf over all possible collision times t and all possiblevalues of vi . If we assume that the initial velocities are randomly distributed over allpossible values, we see that the average value of vi is zero. The term (q E/me)t is thevelocity change of the electron due to the electric eld during one trip between atoms.The average value of the second term of Equation 27.13 is (q E/me ) , where is theaverage time inter val between successive collisions. Because the average value of vf is equal tothe drift velocity, we haveDrift velocity in terms ofmicroscopic quantitiesvfqEmevd(27.14)We can relate this expression for drift velocity to the current in the conductor.Substituting Equation 27.14 into Equation 27.6, we nd that the magnitude of thecurrent density isJCurrent density in terms ofmicroscopic quantitiesnq 2Emenqvd(27.15)where n is the number of charge carriers per unit volume. Comparing this expressionwith Ohms law, JE , we obtain the following relationships for conductivity andresistivity of a conductor:nq 2meConductivity in terms of microscopic quantities(27.16)1Resistivity in terms of microscopic quantitiesmenq 2(27.17)According to this classical model, conductivity and resistivity do not depend on thestrength of the electric eld. This feature is characteristic of a conductor obeyingOhms law.The average time interval between collisions is related to the average distancebetween collisions (that is, the mean free path; see Section 21.7) and the average speedv through the expression(27.18)vExample 27.5Electron Collisions in a Wire(A) Using the data and results from Example 27.1 and theclassical model of electron conduction, estimate the averagetime interval between collisions for electrons in householdcopper wiring.Solution From Equation 27.17, we see thatmenq 2wheresity is n1.7 10 8m for copper and the carrier den8.49 1028 electrons/m3 for the wire describedin Example 27.1. Substitution of these values into theexpression above gives9.11(8.492.510281014m3)(1.610 31 kg10 19 C)2 (1.7108m)s(B) Assuming that the average speed for free electrons incopper is 1.6 106 m/s and using the result from part (A),calculate the mean free path for electrons in copper.S E C T I O N 2 7. 4 Resistance and TemperatureSolution From Equation 27.18,v(1.64.010106 m/s)(2.581014843which is equivalent to 40 nm (compared with atomicspacings of about 0.2 nm). Thus, although the time intervalbetween collisions is very short, an electron in the wiretravels about 200 atomic spacings between collisions.s)m27.4 Resistance and TemperatureOver a limited temperature range, the resistivity of a conductor varies approximatelylinearly with temperature according to the expression0[1(TT0)](27.19)Variation ofwith temperaturewhere is the resistivity at some temperature T (in degrees Celsius), 0 is the resistivityat some reference temperature T0 (usually taken to be 20C), and is the temperature coefcient of resistivity. From Equation 27.19, we see that the temperaturecoefcient of resistivity can be expressed as10 T(27.20)Temperature coefcient ofresistivitywhereT T0.0 is the change in resistivity in the temperature interval TThe temperature coefcients of resistivity for various materials are given inTable 27.1. Note that the unit for is degrees Celsius 1 [(C) 1]. Because resistanceis proportional to resistivity (Eq. 27.11), we can write the variation of resistance asRR 0[1(TT0)](27.21)Use of this property enables us to make precise temperature measurements, as shownin Example 27.6.Quick Quiz 27.6 When does a lightbulb carry more current: (a) just after itis turned on and the glow of the metal lament is increasing, or (b) after it has beenon for a few milliseconds and the glow is steady?Example 27.6A Platinum Resistance ThermometerA resistance thermometer, which measures temperature bymeasuring the change in resistance of a conductor, is madefrom platinum and has a resistance of 50.0at 20.0C.When immersed in a vessel containing melting indium, itsresistance increases to 76.8 . Calculate the melting pointof the indium.TRR0R0[3.9276.8103(50.0C ) 1](50.0)137 CBecause T020.0C, we nd that T, the temperature of themelting indium sample, is 157 C.Solution Solving Equation 27.21 for T and using thevalue for platinum given in Table 27.1, we obtainFor metals like copper, resistivity is nearly proportional to temperature, as shownin Figure 27.10. However, a nonlinear region always exists at very low temperatures,and the resistivity usually reaches some nite value as the temperature approachesabsolute zero. This residual resistivity near absolute zero is caused primarily by theC HAPTE R 2 7 Current and Resistance8440T00collision of electrons with impurities and imper fections in the metal. In contrast, hightemperature resistivity (the linear region) is predominantly characterized by collisionsbetween electrons and metal atoms.Notice that three of the values in Table 27.1 are negative; this indicates that theresistivity of these materials decreases with increasing temperature (Fig. 27.11), whichis indicative of a class of materials called semiconductors. This behavior is due to anincrease in the density of charge carriers at higher temperatures.Because the charge carriers in a semiconductor are often associated with impurityatoms, the resistivity of these materials is very sensitive to the type and concentration ofsuch impurities. We shall return to the study of semiconductors in Chapter 43.27.5TFigure 27.10 Resistivity versustemperature for a metal such ascopper. The curve is linear over awide range of temperatures, andincreases with increasingtemperature. As T approachesabsolute zero (inset), the resistivityapproaches a nite value 0.SuperconductorsThere is a class of metals and compounds whose resistance decreases to zero when theyare below a certain temperature Tc , known as the critical temperature. Thesematerials are known as superconductors. The resistancetemperature graph for asuperconductor follows that of a normal metal at temperatures above Tc (Fig. 27.12).When the temperature is at or below Tc , the resistivity drops suddenly to zero. Thisphenomenon was discovered in 1911 by the Dutch physicist Heike KamerlinghOnnes(18531926) as he worked with mercury, which is a superconductor below 4.2 K.Recent measurements have shown that the resistivities of superconductors below theirmaround 1017 times smaller than the resistivityTc values are less than 4 10 25of copper and in practice considered to be zero.Today thousands of superconductors are known, and as Table 27.3 illustrates, thecritical temperatures of recently discovered superconductors are substantially higherthan initially thought possible. Two kinds of superconductors are recognized. Themore recently identied ones are essentially ceramics with high critical temperatures,whereas superconducting materials such as those observed by KamerlinghOnnes aremetals. If a roomtemperature superconductor is ever identied, its impact ontechnology could be tremendous.The value of Tc is sensitive to chemical composition, pressure, and molecularstructure. It is interesting to note that copper, silver, and gold, which are excellentconductors, do not exhibit superconductivity.TFigure 27.11 Resistivity versustemperature for a puresemiconductor, such as silicon orgermanium.R()0.15Table 27.3Critical Temperatures forVarious Superconductors0.10Material0.05Tc0.004.04.2T (K)4.4Figure 27.12 Resistance versus temperature for asample of mercury (Hg). The graph follows that of anormal metal above the critical temperature Tc . Theresistance drops to zero at Tc , which is 4.2 K for mercury.Tc(K)HgBa2Ca2Cu3O8TlBaCaCuOBiSrCaCuOYBa2Cu3O7Nb3GeNb3SnNbPbHgSnAlZn1341251059223.218.059.467.184.153.721.190.88One of the truly remarkable features of superconductors is that once a current isset up in them, it persists without any applied potential difference (because R 0). Steadycurrents have been observed to persist in superconducting loops for several years withno apparent decay!An important and useful application of superconductivity is in the development ofsuperconducting magnets, in which the magnitudes of the magnetic eld are aboutten times greater than those produced by the best normal electromagnets. Suchsuperconducting magnets are being considered as a means of storing energy.Superconducting magnets are currently used in medical magnetic resonance imaging(MRI) units, which produce highquality images of internal organs without the needfor excessive exposure of patients to xrays or other harmful radiation.For further information on superconductivity, see Section 43.8.27.6Electrical PowerIf a battery is used to establish an electric current in a conductor, there is a continuoustransformation of chemical energy in the battery to kinetic energy of the electrons tointernal energy in the conductor, resulting in an increase in the temperature of theconductor.In typical electric circuits, energy is transferred from a source such as a battery, tosome device, such as a lightbulb or a radio receiver. Let us determine an expressionthat will allow us to calculate the rate of this energy transfer. First, consider the simplecircuit in Figure 27.13, where we imagine energy is being delivered to a resistor.(Resistors are designated by the circuit symbol.) Because the connectingwires also have resistance, some energy is delivered to the wires and some energy to theresistor. Unless noted otherwise, we shall assume that the resistance of the wires is sosmall compared to the resistance of the circuit element that we ignore the energydelivered to the wires.Imagine following a positive quantity of charge Q that is moving clockwise aroundthe circuit in Figure 27.13 from point a through the battery and resistor back to point a.We identify the entire circuit as our system. As the charge moves from a to b throughthe battery, the electric potential energy of the system increases by an amount Q V whilethe chemical potential energy in the battery decreases by the same amount. (Recall fromEq. 25.9 that U q V.) However, as the charge moves from c to d through theresistor, the system loses this electric potential energy during collisions of electrons withatoms in the resistor. In this process, the energy is transformed to internal energycorresponding to increased vibrational motion of the atoms in the resistor. Because wehave neglected the resistance of the interconnecting wires, no energy transformationoccurs for paths bc and da. When the charge returns to point a, the net result is thatsome of the chemical energy in the battery has been delivered to the resistor andresides in the resistor as internal energy associated with molecular vibration.The resistor is normally in contact with air, so its increased temperature will resultin a transfer of energy by heat into the air. In addition, the resistor emits thermal845Courtesy of IBM Research LaboratoryS E C T I O N 2 7. 6 Electrical PowerA small permanent magnetlevitated above a disk of thesuperconductor YBa2Cu3O7, whichis at 77 K.LPITFALL PR EVE NTI O N27.6 MisconceptionsAbout CurrentThere are several commonmisconceptions associated withcurrent in a circuit like that inFigure 27.13. One is that currentcomes out of one terminal of thebattery and is then used up as itpasses through the resistor,leaving current in only one partof the circuit. The truth is thatthe current is the same ever ywherein the circuit. A related misconception has the current comingout of the resistor being smallerthan that going in, because someof the current is used up.Anothermisconceptionhascurrent coming out of bothterminals of the battery, inopposite directions, and thenclashing in the resistor, delivering the energy in this manner.This is not the casethe chargesow in the same rotational senseat all points in the circuit.Ib+acVRdActive Figure 27.13 A circuit consisting of a resistor ofresistance R and a battery having a potential difference V acrossits terminals. Positive charge ows in the clockwise direction.At the Active Figures linkat http://www.pse6.com, youcan adjust the battery voltageand the resistance to see theresulting current in the circuitand power delivered to theresistor.846LC HAPTE R 2 7 Current and ResistancePITFALL PR EVE NTI O N27.7 Charges Do NotMove All the WayAround a Circuit in aShort TimeDue to the very small magnitudeof the drift velocity, it might takehours for a single electron tomake one complete trip aroundthe circuit. In terms of understanding the energy transfer in acircuit, however, it is useful toimagine a charge moving all theway around the circuit.radiation, representing another means of escape for the energy. After some timeinterval has passed, the resistor reaches a constant temperature, at which time theinput of energy from the battery is balanced by the output of energy by heat andradiation. Some electrical devices include heat sinks 4 connected to parts of the circuitto prevent these parts from reaching dangerously high temperatures. These are piecesof metal with many ns. The high thermal conductivity of the metal provides a rapidtransfer of energy by heat away from the hot component, while the large number ofns provides a large sur face area in contact with the air, so that energy can transfer byradiation and into the air by heat at a high rate.Let us consider now the rate at which the system loses electric potential energy asthe charge Q passes through the resistor:dUdtd(Q V )dtdQVdtI Vwhere I is the current in the circuit. The system regains this potential energy when thecharge passes through the battery, at the expense of chemical energy in the battery.The rate at which the system loses potential energy as the charge passes through theresistor is equal to the rate at which the system gains internal energy in the resistor.Thus, the power , representing the rate at which energy is delivered to the resistor, isIVPower delivered to a device(27.22)We derived this result by considering a battery delivering energy to a resistor. However,Equation 27.22 can be used to calculate the power delivered by a voltage source to anydevice carrying a current I and having a potential difference V between its terminals.Using Equation 27.22 and the fact that V I R for a resistor, we can express thepower delivered to the resistor in the alternative formsI 2RPower delivered to a resistorLPITFALL PR EVE NTI O N27.8 Energy Is NotDissipatedIn some books, you may seeEquation 27.23 described as thepower dissipated in a resistor,suggesting that energy disappears.Instead we say energy is deliveredto a resistor. The notion ofdissipation arises because a warmresistor will expel energy byradiation and heat, so that energydelivered by the battery leaves thecircuit. (It does not disappear!)( V )2R(27.23)When I is expressed in amperes, V in volts, and R in ohms, the SI unit of power is thewatt, as it was in Chapter 7 in our discussion of mechanical power. The process bywhich power is lost as internal energy in a conductor of resistance R is often called jouleheating 5; this transformation is also often referred to as an I 2R loss.When transporting energy by electricity through power lines, such as thoseshown in the opening photograph for this chapter, we cannot make the simplifyingassumption that the lines have zero resistance. Real power lines do indeed haveresistance, and power is delivered to the resistance of these wires. Utility companiesseek to minimize the power transformed to internal energy in the lines and maximize the energy delivered to the consumer. BecauseI V, the same amount ofpower can be transported either at high currents and low potential differences or atlow currents and high potential differences. Utility companies choose to transportenergy at low currents and high potential differences primarily for economicreasons. Copper wire is ver y expensive, and so it is cheaper to use highresistancewire (that is, wire having a small crosssectional area; see Eq. 27.11). Thus, in theexpression for the power delivered to a resistor,I 2R , the resistance of the wireis xed at a relatively high value for economic considerations. The I 2R loss can bereduced by keeping the current I as low as possible, which means transferring theenergy at a high voltage. In some instances, power is transported at potentialdifferences as great as 765 kV. Once the electricity reaches your city, the potentialdifference is usually reduced to 4 kV by a device called a transformer. Another4This is another misuse of the word heat that is ingrained in our common language.5It is commonly called joule heating even though the process of heat does not occur. This is anotherexample of incorrect usage of the word heat that has become entrenched in our language.S E C T I O N 2 7. 6 Electrical Power847transformer drops the potential difference to 240 V before the electricity nallyreaches your home. Of course, each time the potential difference decreases, thecurrent increases by the same factor, and the power remains the same. We shalldiscuss transformers in greater detail in Chapter 33.Demands on our dwindling energy supplies have made it necessar y for us to beaware of the energy requirements of our electrical devices. Ever y electricalappliance carries a label that contains the information you need to calculate theappliances power requirements. In many cases, the power consumption in watts isstated directly, as it is on a lightbulb. In other cases, the amount of current used bythe device and the potential difference at which it operates are given. Thisinformation and Equation 27.22 are sufcient for calculating the power requirement of any electrical device.Quick Quiz 27.7 The same potential difference is applied to the twolightbulbs shown in Figure 27.14. Which one of the following statements is true?(a) The 30W bulb carries the greater current and has the higher resistance.(b) The 30W bulb carries the greater current, but the 60W bulb has the higherresistance. (c) The 30W bulb has the higher resistance, but the 60W bulb carriesthe greater current. (d) The 60W bulb carries the greater current and has thehigher resistance.30 WefGeorge Semple60 WcdFigure 27.14 (Quick Quiz 27.7) These lightbulbs operate at their rated power onlywhen they are connected to a 120V source.aQuick Quiz 27.8For the two lightbulbs shown in Figure 27.15, rank thecurrent values at points a through f, from greatest to least.Example 27.7Solution Because VFigure 27.15 (Quick Quiz 27.8)Two lightbulbs connected acrossthe same potential difference.VR120 V8.00(15.0 A)2(8.0015.0 A)1.801.80 kWWhat If? What if the heater were accidentally connected to a240V supply? (This is difcult to do because the shape andorientation of the metal contacts in 240V plugs are differentfrom those in 120V plugs.) How would this affect the currentcarried by the heater and the power rating of the heater?IR , we haveWe can nd the power rating using the expressionI 2R :I 2RbPower in an Electric HeaterAn electric heater is constructed by applying a potentialdifference of 120 V to a Nichrome wire that has a totalresistance of 8.00 . Find the current carried by the wireand the power rating of the heater.IV103 WAnswer If we doubled the applied potential difference,Equation 27.8 tells us that the current would double.According to Equation 27.23,( V )2/R, the powerwould be four times larger.848C HAPTE R 2 7 Current and ResistanceLinking Electricity and ThermodynamicsInteractive(A) What is the required resistance of an immersion heaterthat will increase the temperature of 1.50 kg of water from10.0C to 50.0C in 10.0 min while operating at 110 V?the electrical power. The amount of energy transfer by heatnecessary to raise the temperature of the water is given byEquation 20.4, Q mc T. Thus,Example 27.8(B) Estimate the cost of heating the water.Solution This example allows us to link our new understanding of power in electricity with our experience withspecic heat in thermodynamics (Chapter 20). An immersion heater is a resistor that is inserted into a container ofwater. As energy is delivered to the immersion heater,raising its temperature, energy leaves the sur face of theresistor by heat, going into the water. When the immersionheater reaches a constant temperature, the rate of energydelivered to the resistance by electrical transmission is equalto the rate of energy delivered by heat to the water.(A) To simplify the analysis, we ignore the initial periodduring which the temperature of the resistor increases, andalso ignore any variation of resistance with temperature. Thus,we imagine a constant rate of energy transfer for the entire10.0 min. Setting the rate of energy delivered to the resistorequal to the rate of energy entering the water by heat, we have(V )2RQt(V )2Rmc Tt9:(V )2 tmc TRSubstituting the values given in the statement of theproblem, we haveR(110 V )2(600 s)(1.50 kg)(4186 J/kg C)(50.0 C10.0 C)28.9(B) Because the energy transferred equals power multipliedby time interval, the amount of energy transferred ist(V )2tR(110 V )2(10.0 min)28.969.8 W h1h60.0 min0.069 8 kW hIf the energy is purchased at an estimated price of 10.0 perkilowatthour, the cost isCost(0.069 8 kWh)($0.100/kWh)where Q represents an amount of energy transfer by heatinto the water and we have used Equation 27.23 to express$0.006 980.7 At the Interactive Worked Example link at http://www.pse6.com, you can explore the heating of the water.Example 27.9Current in an Electron BeamIn a certain particle accelerator, electrons emerge withan energy of 40.0 MeV (1 MeV 1.60 10 13 J). Theelectrons emerge not in a steady stream but rather in pulsesat the rate of 250 pulses/s. This corresponds to a timeinterval between pulses of 4.00 ms (Fig. 27.16). Each pulsehas a duration of 200 ns, and the electrons in the pulseconstitute a current of 250 mA. The current is zero betweenpulses.Solution We use Equation 27.2 in the form dQ I dt andintegrate to nd the charge per pulse. While the pulse is on,the current is constant; thus,(A) How many electrons are delivered by the acceleratorper pulse?Dividing this quantity of charge per pulse by the electroniccharge gives the number of electrons per pulse:Q pulseIdt5.004.00 msI t10(2503A)(200109 s)8C2.00 107 sI10t (s)Figure 27.16 (Example 27.9) Current versus time for a pulsed beam of electrons.Summary5.00 10 8 C/pulse1.60 10 19 C/electronElectrons per pulse1011 electrons/pulse3.13We could also compute this power directly. We assume thateach electron has zero energy before being accelerated.Thus, by denition, each electron must go through apotential difference of 40.0 M V to acquire a nal energy of40.0 MeV. Hence, we have(B) What is the average current per pulse delivered by theaccelerator?Solution Average current is given by Equation 27.1,I avQ / t. Because the time interval between pulses is4.00 ms, and because we know the charge per pulse frompart (A), we obtainQ pulseI avt5.004.00101083Cs12.5 AThis represents only 0.005% of the peak current, which is250 mA.(C) What is the peak power delivered by the electron beam?Solution By denition, power is energy delivered per unittime interval. Thus, the peak power is equal to the energydelivered by a pulse divided by the pulse duration:(1)(3.13pulse energypulse duration107 W(2)peakI peak V(250103A)(40.0106 V )10.0 MWWhat If ? What if the requested quantity in part (C) were theaverage power rather than the peak power?Answer Instead of Equation (1), we would use the timeinterval between pulses rather than the duration of a pulse:avpulse energytime interval between pulses(3.131011 electrons/pulse)(40.0 MeV/electron)4.00 10 3 s/pulse1.60 101 MeV13J500 W1011 electrons/pulse)(40.0 MeV/electron)2.00 10 7 s/pulse1.60 101 MeV1.00peak84913 JInstead of Equation (2), we would use the average currentfound in part (B):avIav V(12.5106A)(40.0106 V )500 WNotice that these two calculations agree with each other andthat the average power is much lower than the peak power.10.0 MWS U M MARYThe electric current I in a conductor is dened asdQdtI(27.2)where dQ is the charge that passes through a cross section of the conductor in a timeinterval dt. The SI unit of current is the ampere (A), where 1 A 1 C/s.The average current in a conductor is related to the motion of the charge carriersthrough the relationshipI avnqvd A(27.4)where n is the density of charge carriers, q is the charge on each carrier, vd is the driftspeed, and A is the crosssectional area of the conductor.The magnitude of the current density J in a conductor is the current perunit area:JIAnqvd(27.5)Take a practice test forthis chapter by clicking onthe Practice Test link athttp://www.pse6.com.850C HAPTE R 2 7 Current and ResistanceThe current density in an ohmic conductor is proportional to the electric eldaccording to the expressionJE(27.7)The proportionality constant is called the conductivity of the material of which theconductor is made. The inverse ofis known as resistivity (that is,1/ ).Equation 27.7 is known as Ohms law, and a material is said to obey this law if the ratioof its current density J to its applied electric eld E is a constant that is independent ofthe applied eld.The resistance R of a conductor is dened asRVI(27.8)where V is the potential difference across it, and I is the current it carries.The SI unit of resistance is volts per ampere, which is dened to be 1 ohm ( );that is, 11 V/A. If the resistance is independent of the applied potentialdifference, the conductor obeys Ohms law.For a uniform block of material of cross sectional area A and length , theresistance over the length isR(27.11)Awhere is the resistivity of the material.In a classical model of electrical conduction in metals, the electrons are treated asmolecules of a gas. In the absence of an electric eld, the average velocity of theelectrons is zero. When an electric eld is applied, the electrons move (on the average)with a drift velocity vd that is opposite the electric eld and given by the expressionvdqEme(27.14)where is the average time inter val between electronatom collisions, me is the massof the electron, and q is its charge. According to this model, the resistivity of themetal ismenq 2(27.17)where n is the number of free electrons per unit volume.The resistivity of a conductor varies approximately linearly with temperatureaccording to the expression0[1(TT0)](27.19)where is the temperature coefcient of resistivity and 0 is the resistivity at somereference temperature T 0.If a potential difference V is maintained across a circuit element, the power, orrate at which energy is supplied to the element, isIVBecause the potential difference across a resistor is given bythe power delivered to a resistor in the formI 2R(V )2R(27.22)VIR, we can express(27.23)The energy delivered to a resistor by electrical transmission appears in the form ofinternal energy in the resistor.Questions851QU ESTIONS1. In an analogy between electric current and automobiletrafc ow, what would correspond to charge? What wouldcorrespond to current?9. When the voltage across a certain conductor is doubled,the current is observed to increase by a factor of three.What can you conclude about the conductor?2. Newspaper articles often contain a statement such as10 000 volts of electricity surged through the victimsbody. What is wrong with this statement?10. In the water analogy of an electric circuit, what corresponds to the power supply, resistor, charge, and potentialdifference?3. What factors affect the resistance of a conductor?11. Use the atomic theory of matter to explain why theresistance of a material should increase as its temperatureincreases.4. What is the difference between resistance and resistivity?5. Two wires A and B of circular cross section are made of thesame metal and have equal lengths, but the resistance ofwire A is three times greater than that of wire B. What isthe ratio of their crosssectional areas? How do their radiicompare?6. Do all conductors obey Ohms law? Give examples tojustify your answer.7. We have seen that an electric eld must exist inside aconductor that carries a current. How is it possible in viewof the fact that in electrostatics we concluded that theelectric eld must be zero inside a conductor?8. A ver y large potential difference is not necessarilyrequired to produce long sparks in air. With a devicecalled Jacobs ladder, a potential difference of about 10 kVproduces an electric arc a few millimeters long betweenthe bottom ends of two cur ved rods that project upwardfrom the power supply. (The device is seen in classicmadscientist horror movies and in Figure Q27.8.) Thearc rises, climbing the rods and getting longer andlonger. It disappears when it reaches the top; then a newspark immediately forms at the bottom and the processrepeats. Explain these phenomena. Why does the arcrise? Why does a new arc appear only after the previousone is gone?12. Why might a good electrical conductor also be a goodthermal conductor?13. How does the resistance for copper and for silicon changewith temperature? Why are the behaviors of these twomaterials different?14. Explain how a current can persist in a superconductorwithout any applied voltage.15. What single experimental requirement makes superconducting devices expensive to operate? In principle, canthis limitation be overcome?16. What would happen to the drift velocity of the electrons ina wire and to the current in the wire if the electrons couldmove freely without resistance through the wire?17. If charges ow very slowly through a metal, why does it notrequire several hours for a light to come on when youthrow a switch?18. In a conductor, changes in the electric eld that drives theelectrons through the conductor propagate with a speedclose to the speed of light, although the drift velocity ofthe electrons is very small. Explain how these statementscan both be true. Does one particular electron move fromone end of the conductor to the other?19. Two conductors of the same length and radius areconnected across the same potential difference. Oneconductor has twice the resistance of the other. To whichconductor is more power delivered?20. Two lightbulbs both operate from 120 V. One has a powerof 25 W and the other 100 W. Which bulb has higherresistance? Which bulb carries more current?21. Car batteries are often rated in amperehours. Does thisdesignate the amount of current, power, energy, or chargethat can be drawn from the battery?22. If you were to design an electric heater using Nichromewire as the heating element, what parameters of the wirecould you vary to meet a specic power output, such as1 000 W ?Figure Q27.8C HAPTE R 2 7 Current and Resistance852PROBLEMS1, 2, 3straightforward, intermediate, challengingfull solution available in the Student Solutions Manual and Study Guidecoached solution with hints available at http://www.pse6.comcomputer useful in solving problempaired numerical and symbolic problemsSection 27.1Electric Currentradius of cross section A1 is 0.400 cm. (a) What is themagnitude of the current density across A1? (b) If thecurrent density across A2 is onefourth the value across A1,what is the radius of the conductor at A2?1. In a particular cathode ray tube, the measured beamcurrent is 30.0 A. How many electrons strike the tubescreen every 40.0 s?2. A teapot with a sur face area of 700 cm2 is to be silverplated. It is attached to the negative electrode of anelectrolytic cell containing silver nitrate (Ag NO3 ). Ifthe cell is powered by a 12.0V battery and has a resistanceof 1.80 , how long does it take for a 0.133mm layer ofsilver to build up on the teapot? (The density of silver is10.5 103 kg/m3.)3.Suppose that the current through a conductordecreases exponentially with time according to theequation I(t ) I0e t/ where I0 is the initial current (att0), and is a constant having dimensions of time.Consider a xed observation point within the conductor.(a) How much charge passes this point between t 0 andt? (b) How much charge passes this point betweent 0 and t 10 ? (c) What If ? How much charge passesthis point between t 0 and t?4. In the Bohr model of the hydrogen atom, an electronin the lowest energy state follows a circular path5.29 10 11 m from the proton. (a) Show that the speedof the electron is 2.19 106 m/s. (b) What is the effectivecurrent associated with this orbiting electron?5. A small sphere that carries a charge q is whirled in a circleat the end of an insulating string. The angular frequencyof rotation is . What average current does this rotatingcharge represent?6. The quantity of charge q (in coulombs) that has passedthrough a sur face of area 2.00 cm2 varies with timeaccording to the equation q 4t 3 5t 6, where t is inseconds. (a) What is the instantaneous current throughthe sur face at t 1.00 s? (b) What is the value of thecurrent density?7. An electric current is given by the expression I (t )100 sin(120 t), where I is in amperes and t is in seconds.What is the total charge carried by the current from t 0to t (1/240) s?8. Figure P27.8 represents a section of a circular conductorof nonuniform diameter carrying a current of 5.00 A. TheA2A1IFigure P27.89. The electron beam emerging from a certain highenergyelectron accelerator has a circular cross section of radius1.00 mm. (a) The beam current is 8.00 A. Find the currentdensity in the beam, assuming that it is uniform throughout.(b) The speed of the electrons is so close to the speed oflight that their speed can be taken as c3.00 108 m/swith negligible error. Find the electron density in the beam.(c) How long does it take for Avogadros number ofelectrons to emerge from the accelerator?10. A Van de Graaff generator produces a beam of 2.00MeVdeuterons, which are heavy hydrogen nuclei containing aproton and a neutron. (a) If the beam current is 10.0 A,how far apart are the deuterons? (b) Is the electric forceof repulsion among them a signicant factor in beamstability? Explain.11. An aluminum wire having a crosssectional area of4.00 10 6 m2 carries a current of 5.00 A. Find the driftspeed of the electrons in the wire. The density ofaluminum is 2.70 g/cm3. Assume that one conductionelectron is supplied by each atom.Section 27.2 Resistance12. Calculate the current density in a gold wire at 20C, if anelectric eld of 0.740 V/m exists in the wire.13. A lightbulb has a resistance of 240 when operating witha potential difference of 120 V across it. What is thecurrent in the lightbulb?14. A resistor is constructed of a carbon rod that has auniform crosssectional area of 5.00 mm2. When apotential difference of 15.0 V is applied across the ends ofthe rod, the rod carries a current of 4.00 10 3 A. Find(a) the resistance of the rod and (b) the rods length.15.A 0.900V potential difference is maintained across a1.50m length of tungsten wire that has a crosssectionalarea of 0.600 mm2. What is the current in the wire?16. A conductor of uniform radius 1.20 cm carries a currentof 3.00 A produced by an electric eld of 120 V/m. Whatis the resistivity of the material?17. Suppose that you wish to fabricate a uniform wire out of1.00 g of copper. If the wire is to have a resistance ofR0.500 , and if all of the copper is to be used, whatwill be (a) the length and (b) the diameter of this wire?18. Gold is the most ductile of all metals. For example, onegram of gold can be drawn into a wire 2.40 km long. Whatis the resistance of such a wire at 20C? You can nd thenecessary reference information in this textbook.Problems19. (a) Make an orderofmagnitude estimate of the resistance between the ends of a rubber band. (b) Make anorderofmagnitude estimate of the resistance betweenthe heads and tails sides of a penny. In each case statewhat quantities you take as data and the values youmeasure or estimate for them. (c) WARNING! Do not trythis at home! What is the order of magnitude of thecurrent that each would carry if it were connected acrossa 120V power supply?20. A solid cube of silver (density 10.5 g/cm3) has a mass of90.0 g. (a) What is the resistance between opposite faces ofthe cube? (b) Assume each silver atom contributes oneconduction electron. Find the average drift speed ofelectrons when a potential difference of 1.00 10 5 V isapplied to opposite faces. The atomic number of silver is47, and its molar mass is 107.87 g/mol.21. A metal wire of resistance R is cut into three equal piecesthat are then connected side by side to form a new wirethe length of which is equal to onethird the originallength. What is the resistance of this new wire?22. Aluminum and copper wires of equal length are found tohave the same resistance. What is the ratio of their radii?23. A current density of 6.00 10 13 A/m2 exists in the atmosphere at a location where the electric eld is 100 V/m.Calculate the electrical conductivity of the Earths atmosphere in this region.24. The rod in Figure P27.24 is made of two materials. The gure is not drawn to scale. Each conductor has a squarecross section 3.00 mm on a side. The rst material has aresistivity of 4.00 10 3m and is 25.0 cm long, whilethe second material has a resistivity of 6.00 10 3mand is 40.0 cm long. What is the resistance between theends of the rod?40.0 cm25.0 cmFigure P27.24Section 27.3 A Model for Electrical Conduction25.If the magnitude of the drift velocity of free electronsin a copper wire is 7.84 10 4 m/s, what is the electriceld in the conductor?26. If the current carried by a conductor is doubled, whathappens to the (a) charge carrier density? (b) currentdensity? (c) electron drift velocity? (d) average timeinterval between collisions?27. Use data from Example 27.1 to calculate the collisionmean free path of electrons in copper. Assume the averagethermal speed of conduction electrons is 8.60 105 m/s.Section 27.4 Resistance and Temperature28. While taking photographs in Death Valley on a day whenthe temperature is 58.0C, Bill Hiker nds that a certainvoltage applied to a copper wire produces a current of1.000 A. Bill then travels to Antarctica and applies the853same voltage to the same wire. What current does heregister there if the temperature is 88.0C? Assume thatno change occurs in the wires shape and size.29. A certain lightbulb has a tungsten lament with aresistance of 19.0when cold and 140when hot.Assume that the resistivity of tungsten varies linearly withtemperature even over the large temperature rangeinvolved here, and nd the temperature of the hotlament. Assume the initial temperature is 20.0C.30. A carbon wire and a Nichrome wire are connected inseries, so that the same current exists in both wires. If thecombination has a resistance of 10.0 k at 0C, what is theresistance of each wire at 0C so that the resistance ofthe combination does not change with temperature? Thetotal or equivalent resistance of resistors in series isthe sum of their individual resistances.31. An aluminum wire with a diameter of 0.100 mm has auniform electric eld of 0.200 V/m imposed along itsentire length. The temperature of the wire is 50.0C.Assume one free electron per atom. (a) Use the information in Table 27.1 and determine the resistivity. (b) What isthe current density in the wire? (c) What is the totalcurrent in the wire? (d) What is the drift speed of theconduction electrons? (e) What potential difference mustexist between the ends of a 2.00m length of the wire toproduce the stated electric eld?32. Review problem. An aluminum rod has a resistance of1.234at 20.0C. Calculate the resistance of the rod at120C by accounting for the changes in both the resistivityand the dimensions of the rod.33. What is the fractional change in the resistance of an ironlament when its temperature changes from 25.0C to50.0C?34. The resistance of a platinum wire is to be calibrated forlowtemperature measurements. A platinum wire withresistance 1.00 at 20.0C is immersed in liquid nitrogenat 77 K ( 196C). If the temperature response of the platinum wire is linear, what is the expected resistance of theplatinum wire at196C? ( platinum 3.92 10 3/C)35. The temperature of a sample of tungsten is raised while asample of copper is maintained at 20.0C. At what temperature will the resistivity of the tungsten be four times thatof the copper?Section 27.6 Electrical Power36. A toaster is rated at 600 W when connected to a 120Vsource. What current does the toaster carry, and what is itsresistance?37. A Van de Graaff generator (see Figure 25.29) is operatingso that the potential difference between the highvoltageelectrode B and the charging needles at A is 15.0 kV.Calculate the power required to drive the belt againstelectrical forces at an instant when the effective currentdelivered to the highvoltage electrode is 500 A.38. In a hydroelectric installation, a turbine delivers 1 500 hpto a generator, which in turn transfers 80.0% of themechanical energy out by electrical transmission. Under854C HAPTE R 2 7 Current and Resistancethese conditions, what current does the generator deliverat a terminal potential difference of 2 000 V ?39.What is the required resistance of an immersion heaterthat increases the temperature of 1.50 kg of water from10.0C to 50.0C in 10.0 min while operating at 110 V ?40. One rechargeable battery of mass 15.0 g delivers to a CDplayer an average current of 18.0 mA at 1.60 V for 2.40 hbefore the battery needs to be recharged. The rechargermaintains a potential difference of 2.30 V across the batteryand delivers a charging current of 13.5 mA for 4.20 h.(a) What is the efciency of the battery as an energy storagedevice? (b) How much internal energy is produced in thebattery during one chargedischarge cycle? (b) If the batteryis surrounded by ideal thermal insulation and has an overalleffective specic heat of 975 J/kgC, by how much will itstemperature increase during the cycle?41. Suppose that a voltage surge produces 140 V for amoment. By what percentage does the power output of a120V, 100W lightbulb increase? Assume that its resistancedoes not change.42. A 500W heating coil designed to operate from 110 V ismade of Nichrome wire 0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains constant atits 20.0C value, nd the length of wire used. (b) What If ?Now consider the variation of resistivity with temperature.What power will the coil of part (a) actually deliver when itis heated to 1 200C?43. A coil of Nichrome wire is 25.0 m long. The wire has adiameter of 0.400 mm and is at 20.0C. If it carries acurrent of 0.500 A, what are (a) the magnitude of theelectric eld in the wire, and (b) the power delivered to it?(c) What If ? If the temperature is increased to 340C andthe voltage across the wire remains constant, what is thepower delivered?44. Batteries are rated in terms of amperehours (A h). Forexample, a battery that can produce a current of 2.00 Afor 3.00 h is rated at 6.00 A h. (a) What is the totalenergy, in kilowatthours, stored in a 12.0V battery ratedat 55.0 A h? (b) At $0.060 0 per kilowatthour, what is thevalue of the electricity produced by this battery?45. A 10.0V battery is connected to a 120 resistor. Ignoringthe internal resistance of the battery, calculate the powerdelivered to the resistor.46. Residential building codes typically require the use of12gauge copper wire (diameter 0.205 3 cm) for wiringreceptacles. Such circuits carry currents as large as 20 A.A wire of smaller diameter (with a higher gauge number)could carry this much current, but the wire could rise to ahigh temperature and cause a re. (a) Calculate the rateat which internal energy is produced in 1.00 m of 12gaugecopper wire carrying a current of 20.0 A. (b) What If ?Repeat the calculation for an aluminum wire. Would a12gauge aluminum wire be as safe as a copper wire?47. An 11.0W energyefcient uorescent lamp is designed toproduce the same illumination as a conventional 40.0Wincandescent lightbulb. How much money does the userof the energyefcient lamp save during 100 hours of use?Assume a cost of $0.080 0/kWh for energy from the powercompany.48. We estimate that 270 million plugin electric clocks are inthe United States, approximately one clock for eachperson. The clocks convert energy at the average rate2.50 W. To supply this energy, how many metric tons ofcoal are burned per hour in coalred electric generatingplants that are, on average, 25.0% efcient? The heat ofcombustion for coal is 33.0 MJ/kg.49. Compute the cost per day of operating a lamp that draws acurrent of 1.70 A from a 110V line. Assume the cost ofenergy from the power company is $0.060 0/kWh.50. Review problem. The heating element of a coffee makeroperates at 120 V and carries a current of 2.00 A.Assuming that the water absorbs all of the energy delivered to the resistor, calculate how long it takes to raise thetemperature of 0.500 kg of water from room temperature(23.0C) to the boiling point.51. A certain toaster has a heating element made of Nichromewire. When the toaster is rst connected to a 120V source(and the wire is at a temperature of 20.0C), the initialcurrent is 1.80 A. However, the current begins to decreaseas the heating element warms up. When the toasterreaches its nal operating temperature, the current dropsto 1.53 A. (a) Find the power delivered to the toaster whenit is at its operating temperature. (b) What is the naltemperature of the heating element?52. The cost of electricity varies widely through the UnitedStates; $0.120/kWh is one typical value. At this unitprice, calculate the cost of (a) leaving a 40.0W porchlight on for two weeks while you are on vacation,(b) making a piece of dark toast in 3.00 min with a970W toaster, and (c) dr ying a load of clothes in40.0 min in a 5 200W dr yer.53. Make an orderofmagnitude estimate of the cost of onepersons routine use of a hair dryer for 1 yr. If you donot use a blow dryer yourself, observe or interviewsomeone who does. State the quantities you estimate andtheir values.Additional Problems54. One lightbulb is marked 25 W 120 V, and another 100 W120 V; this means that each bulb has its respective powerdelivered to it when plugged into a constant 120Vpotential difference. (a) Find the resistance of each bulb.(b) How long does it take for 1.00 C to pass through thedim bulb? Is the charge different in any way upon its exitfrom the bulb versus its entry? (c) How long does it takefor 1.00 J to pass through the dim bulb? By what mechanisms does this energy enter and exit the bulb? (d) Findhow much it costs to run the dim bulb continuously for30.0 days if the electric company sells its product at$0.070 0 per kW h. What product does the electric companysell? What is its price for one SI unit of this quantity?55. A charge Q is placed on a capacitor of capacitance C. Thecapacitor is connected into the circuit shown in FigureP27.55, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is thenclosed and the circuit comes to equilibrium. In terms of Qand C, nd (a) the nal potential difference between theplates of each capacitor, (b) the charge on each capacitor,Problemsand (c) the nal energy stored in each capacitor. (d) Findthe internal energy appearing in the resistor.CRFigure P27.5556. A highvoltage transmission line with a diameter of 2.00 cmand a length of 200 km carries a steady current of 1 000 A.If the conductor is copper wire with a free charge densityof 8.49 1028 electrons/m3, how long does it take oneelectron to travel the full length of the line?57. A more general denition of the temperature coefcientof resistivity is1ddTwhere is the resistivity at temperature T. (a) Assumingthat is constant, show that(T T0)where 0 is the resistivity at temperature T0. (b) Usingthe series expansion e x 1 x for x1, show thatthe resistivity is given approximately by the expression(T T0)] for (T T0)1.0[158. A highvoltage transmission line carries 1 000 A starting at700 kV for a distance of 100 mi. If the resistance in thewire is 0.500 /mi, what is the power loss due to resistivelosses?59.An experiment is conducted to measure theelectrical resistivity of Nichrome in the form of wires withdifferent lengths and crosssectional areas. For one set ofmeasurements, a student uses 30gauge wire, which has acrosssectional area of 7.30 10 8 m2. The studentmeasures the potential difference across the wire and thecurrent in the wire with a voltmeter and an ammeter,respectively. For each of the measurements given in thetable taken on wires of three different lengths, calculatethe resistance of the wires and the corresponding values ofthe resistivity. What is the average value of the resistivity,and how does this value compare with the value given inTable 27.1?L (m)0.5401.0281.543V (V)5.225.825.94I (A)R( )house for a load current of 110 A. For this load current,nd (b) the power the customer is receiving and (c) theelectric power lost in the copper wires.61. A straight cylindrical wire lying along the x axis has alength of 0.500 m and a diameter of 0.200 mm. It is madeof a material that obeys Ohms law with a resistivity of4.00 10 8m. Assume that a potential of 4.00 V ismaintained at x 0, and that V 0 at x 0.500 m. Find(a) the electric eld E in the wire, (b) the resistance of thewire, (c) the electric current in the wire, and (d) thecurrent density J in the wire. Express vectors in vectornotation. (e) Show that EJ.3C0e855(m)0.5000.2760.18760. An electric utility company supplies a customers housefrom the main power lines (120 V) with two copper wires,each of which is 50.0 m long and has a resistance of0.108 per 300 m. (a) Find the voltage at the customers62. A straight cylindrical wire lying along the x axis has alength L and a diameter d. It is made of a material thatobeys Ohms law with a resistivity . Assume that potentialV is maintained at x 0, and that the potential is zero atx L . In terms of L, d, V, , and physical constants, deriveexpressions for (a) the electric eld in the wire, (b) theresistance of the wire, (c) the electric current in the wire,and (d) the current density in the wire. Express vectors invector notation. (e) Prove that EJ.63. The potential difference across the lament of a lamp ismaintained at a constant level while equilibrium temperature is being reached. It is observed that the steadystatecurrent in the lamp is only one tenth of the current drawnby the lamp when it is rst turned on. If the temperaturecoefcient of resistivity for the lamp at 20.0C is0.004 50 (C) 1, and if the resistance increases linearlywith increasing temperature, what is the nal operatingtemperature of the lament?64. The current in a resistor decreases by 3.00 A when thevoltage applied across the resistor decreases from 12.0 V to6.00 V. Find the resistance of the resistor.65. An electric car is designed to run off a bank of 12.0V batteries with total energy storage of 2.00 107 J. (a) If theelectric motor draws 8.00 kW, what is the current deliveredto the motor? (b) If the electric motor draws 8.00 kW asthe car moves at a steady speed of 20.0 m/s, how far willthe car travel before it is out of juice?66. Review problem. When a straight wire is heated, itsresistance is given by R R 0[1(T T0)] according toEquation 27.21, where is the temperature coefcient ofresistivity. (a) Show that a more precise result, one thatincludes the fact that the length and area of the wirechange when heated, isRR 0[1(T[1T0)][1(T2 (T T0)]T0)]whereis the coefcient of linear expansion (seeChapter 19). (b) Compare these two results for a2.00mlong copper wire of radius 0.100 mm, rst at20.0C and then heated to 100.0C.67. The temperature coefcients of resistivity in Table 27.1 weredetermined at a temperature of 20C. What would they be at0C? Note that the temperature coefcient of resistivity at20C satises(T T0)], where 0 is the0[1resistivity of the material at T0 20C. The temperatureC HAPTE R 2 7 Current and Resistance856coefcient of resistivityat 0C must satisfy the expressionT ], where 0 is the resistivity of the material0[1at 0C.71. Material with uniform resistivity is formed into a wedgeas shown in Figure P27.71. Show that the resistancebetween face A and face B of this wedge is68. An oceanographer is studying how the ion concentrationin sea water depends on depth. She does this by loweringinto the water a pair of concentric metallic cylinders(Fig. P27.68) at the end of a cable and taking data todetermine the resistance between these electrodes as afunction of depth. The water between the two cylindersforms a cylindrical shell of inner radius ra , outer radius rb ,and length L much larger than rb . The scientist applies apotential difference V between the inner and outersur faces, producing an outward radial current I. Letrepresent the resistivity of the water. (a) Find theresistance of the water between the cylinders in terms of L,, ra , and rb . (b) Express the resistivity of the water interms of the measured quantities L, ra , rb , V, and I.rbLRw( y 2Face By1y2Lw72. A material of resistivity is formed into the shape of atruncated cone of altitude h as shown in Figure P27.72.The bottom end has radius b, and the top end has radiusa. Assume that the current is distributed uniformly overany circular cross section of the cone, so that the currentdensity does not depend on radial position. (The currentdensity does vary with position along the axis of the cone.)Show that the resistance between the two ends is describedby the expression69. In a certain stereo system, each speaker has a resistance of4.00 . The system is rated at 60.0 W in each channel, andeach speaker circuit includes a fuse rated 4.00 A. Is thissystem adequately protected against overload? Explainyour reasoning.70. A close analogy exists between the ow of energy by heatbecause of a temperature difference (see Section 20.7)and the ow of electric charge because of a potentialdifference. The energy dQ and the electric charge dq canboth be transported by free electrons in the conductingmaterial. Consequently, a good electrical conductor isusually a good thermal conductor as well. Consider a thinconducting slab of thickness dx, area A, and electricalconductivity , with a potential difference dV betweenopposite faces. Show that the current I dq/dt is given bythe equation on the left below :AdVdxThermal Conduction(Eq. 20.14)dQdty2y1Figure P27.71Figure P27.68dqdtlnFace AraLCharge Conductiony 1)kAdTdxIn the analogous thermal conduction equation on theright, the rate of energy ow dQ /dt (in SI units of joulesper second) is due to a temperature gradient dT/dx, in amaterial of thermal conductivity k. State analogous rulesrelating the direction of the electric current to the changein potential, and relating the direction of energy ow tothe change in temperature.habRahbFigure P27.7273. The dielectric material between the plates of a parallelplate capacitor always has some nonzero conductivity .Let A represent the area of each plate and d the distancebetween them. Let represent the dielectric constant ofthe material. (a) Show that the resistance R and thecapacitance C of the capacitor are related byRC0(b) Find the resistance between the plates of a 14.0nFcapacitor with a fused quartz dielectric.74.The currentvoltage characteristic curve for a semiconductor diode as a function of temperature T is given bythe equationII 0(e e V/k BT1)Answers to Quick QuizzesHere the rst symbol e represents Eulers number, the baseof natural logarithms. The second e is the charge on theelectron. The k B stands for Boltzmanns constant, and T isthe absolute temperature. Set up a spreadsheet tocalculate I and RV/I for V 0.400 V to 0.600 V inincrements of 0.005 V. Assume I0 1.00 nA. Plot R versusV for T 280 K, 300 K, and 320 K.75. Review problem. A parallelplate capacitor consists ofsquare plates of edge lengththat are separated by adistance d, where d. A potential difference V ismaintained between the plates. A material of dielectricconstant lls half of the space between the plates. Thedielectric slab is now withdrawn from the capacitor, asshown in Figure P27.75. (a) Find the capacitance whenthe left edge of the dielectric is at a distance x from thecenter of the capacitor. (b) If the dielectric is removed at aconstant speed v, what is the current in the circuit as thedielectric is being withdrawn?vVdxFigure P27.75Answers to Quick Quizzes27.1 d, b c, a. The current in part (d) is equivalent to twopositive charges moving to the left. Parts (b) and (c)each represent four positive charges moving in the samedirection because negative charges moving to the left areequivalent to positive charges moving to the right. Thecurrent in part (a) is equivalent to ve positive chargesmoving to the right.27.2 (b). The currents in the two paths add numerically toequal the current coming into the junction, withoutregard for the directions of the two wires coming out of857the junction. This is indicative of scalar addition. Eventhough we can assign a direction to a current, it is not avector. This suggests a deeper meaning for vectorsbesides that of a quantity with magnitude and direction.27.3 (a). The current in each section of the wire is the sameeven though the wire constricts. As the crosssectionalarea A decreases, the drift velocity must increase in orderfor the constant current to be maintained, in accordancewith Equation 27.4. As A decreases, Equation 27.11 tellsus that R increases.27.4 (b). The doubling of the radius causes the area A to befour times as large, so Equation 27.11 tells us that theresistance decreases.27.5 (b). The slope of the tangent to the graph line at a pointis the reciprocal of the resistance at that point. Becausethe slope is increasing, the resistance is decreasing.27.6 (a). When the lament is at room temperature, itsresistance is low, and hence the current is relatively large.As the lament warms up, its resistance increases, andthe current decreases. Older lightbulbs often fail just asthey are turned on because this large initial currentspike produces rapid temperature increase andmechanical stress on the lament, causing it to break.27.7 (c). Because the potential difference V is the sameacross the two bulbs and because the power delivered toa conductor isI V, the 60W bulb, with its higherpower rating, must carry the greater current. The 30Wbulb has the higher resistance because it draws lesscurrent at the same potential difference.27.8 Ia Ib Ic Id Ie If . The current Ia leaves thepositive terminal of the battery and then splits to owthrough the two bulbs; thus, Ia Ic Ie . From QuickQuiz 27.7, we know that the current in the 60W bulb isgreater than that in the 30W bulb. Because charge doesnot build up in the bulbs, we know that the same amountof charge owing into a bulb from the left must ow outon the right; consequently, Ic Id and Ie If . The twocurrents leaving the bulbs recombine to form the currentback into the battery, If Id Ib ....
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 Fall '09
 WilfredNgwa
 Physics, Current, Resistance, Energy, Power

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